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(I know the title isn’t clear. I didn’t know how to best describe my question in the title)

So, we have a position-time relation : $\vec r$ = $2t\hat i$ + $4t^2\hat j$

It shows that the particle is initially at $x$ = $0$, $y$ = $0$

Its corresponding velocity vector is : $\vec v$ = $2\hat i$ + $8t\hat j$.

It means its $v$ along $x-axis$ is constant throughout the motion of the particle.

It’sIts trajectory equation is $y$ = $x^2$, which is a parabola symmetric about $y-axis$ (although we’d only use the part of the curve on the right side of $+ve$ $y-axis$)

From its $y$ vs $x$ curve, I tried to obtain its $v_y$ vs $v_x$ relation, and I got : $v_y$ = $2xv_x$

Now if I put any value of $x$ (particle’s position) and its velocity along $x-axis$ at that value of $x$, I correctly get the value of $v_y$. For example, the particle was at $x$ = $2$ at $t$ = $1$, and its $v_x$ at $t$ = $1$ was $2$ units. Substituting these values I get $v_y$ = $8$ units, which is true, as we can verify from the velocity-time relation that its $v_y$ was $8$ units at $t$ = $1$.

I tried other values of $x$ & $v_x$, or $x$ & $v_y$, and I always seemed to get correct values of $v_y$, and $v_x$. Except for when I tried calculating $v_x$ when the particle was at $x=0$ at $t$ = $0$.

Since $v_x$ = $\frac{v_y}{2x}$, if I put $x$ = $0$ and $v_y$ = $0$, I end up with an indeterminate form $\frac{0}{0}$. ($v_y$ was $0$ at $x$ = $0$)

Shouldn’t the $v_y$ vs $v_x$ relation show that $v_x$ is equal to $2$ units at $x$ = $0$? Actually $v_x$ is constant, and equal to $2$ units at any instant, according to the velocity-time relation.

Am I doing something wrong?

(I know the title isn’t clear. I didn’t know how to best describe my question in the title)

So, we have a position-time relation : $\vec r$ = $2t\hat i$ + $4t^2\hat j$

It shows that the particle is initially at $x$ = $0$, $y$ = $0$

Its corresponding velocity vector is : $\vec v$ = $2\hat i$ + $8t\hat j$.

It means its $v$ along $x-axis$ is constant throughout the motion of the particle.

It’s trajectory equation is $y$ = $x^2$, which is a parabola symmetric about $y-axis$ (although we’d only use the part of the curve on the right side of $+ve$ $y-axis$)

From its $y$ vs $x$ curve, I tried to obtain its $v_y$ vs $v_x$ relation, and I got : $v_y$ = $2xv_x$

Now if I put any value of $x$ (particle’s position) and its velocity along $x-axis$ at that value of $x$, I correctly get the value of $v_y$. For example, the particle was at $x$ = $2$ at $t$ = $1$, and its $v_x$ at $t$ = $1$ was $2$ units. Substituting these values I get $v_y$ = $8$ units, which is true, as we can verify from the velocity-time relation that its $v_y$ was $8$ units at $t$ = $1$.

I tried other values of $x$ & $v_x$, or $x$ & $v_y$, and I always seemed to get correct values of $v_y$, and $v_x$. Except for when I tried calculating $v_x$ when the particle was at $x=0$ at $t$ = $0$.

Since $v_x$ = $\frac{v_y}{2x}$, if I put $x$ = $0$ and $v_y$ = $0$, I end up with an indeterminate form $\frac{0}{0}$. ($v_y$ was $0$ at $x$ = $0$)

Shouldn’t the $v_y$ vs $v_x$ relation show that $v_x$ is equal to $2$ units at $x$ = $0$? Actually $v_x$ is constant, and equal to $2$ units at any instant, according to the velocity-time relation.

Am I doing something wrong?

(I know the title isn’t clear. I didn’t know how to best describe my question in the title)

So, we have a position-time relation : $\vec r$ = $2t\hat i$ + $4t^2\hat j$

It shows that the particle is initially at $x$ = $0$, $y$ = $0$

Its corresponding velocity vector is : $\vec v$ = $2\hat i$ + $8t\hat j$.

It means its $v$ along $x-axis$ is constant throughout the motion of the particle.

Its trajectory equation is $y$ = $x^2$, which is a parabola symmetric about $y-axis$ (although we’d only use the part of the curve on the right side of $+ve$ $y-axis$)

From its $y$ vs $x$ curve, I tried to obtain its $v_y$ vs $v_x$ relation, and I got : $v_y$ = $2xv_x$

Now if I put any value of $x$ (particle’s position) and its velocity along $x-axis$ at that value of $x$, I correctly get the value of $v_y$. For example, the particle was at $x$ = $2$ at $t$ = $1$, and its $v_x$ at $t$ = $1$ was $2$ units. Substituting these values I get $v_y$ = $8$ units, which is true, as we can verify from the velocity-time relation that its $v_y$ was $8$ units at $t$ = $1$.

I tried other values of $x$ & $v_x$, or $x$ & $v_y$, and I always seemed to get correct values of $v_y$, and $v_x$. Except for when I tried calculating $v_x$ when the particle was at $x=0$ at $t$ = $0$.

Since $v_x$ = $\frac{v_y}{2x}$, if I put $x$ = $0$ and $v_y$ = $0$, I end up with an indeterminate form $\frac{0}{0}$. ($v_y$ was $0$ at $x$ = $0$)

Shouldn’t the $v_y$ vs $v_x$ relation show that $v_x$ is equal to $2$ units at $x$ = $0$? Actually $v_x$ is constant, and equal to $2$ units at any instant, according to the velocity-time relation.

Am I doing something wrong?

Source Link
4d_
4d_
  • 876
  • 7
  • 28

$v_y$ vs $v_x$ curve, getting weird value of $v_y$ at $x$ = $0$

(I know the title isn’t clear. I didn’t know how to best describe my question in the title)

So, we have a position-time relation : $\vec r$ = $2t\hat i$ + $4t^2\hat j$

It shows that the particle is initially at $x$ = $0$, $y$ = $0$

Its corresponding velocity vector is : $\vec v$ = $2\hat i$ + $8t\hat j$.

It means its $v$ along $x-axis$ is constant throughout the motion of the particle.

It’s trajectory equation is $y$ = $x^2$, which is a parabola symmetric about $y-axis$ (although we’d only use the part of the curve on the right side of $+ve$ $y-axis$)

From its $y$ vs $x$ curve, I tried to obtain its $v_y$ vs $v_x$ relation, and I got : $v_y$ = $2xv_x$

Now if I put any value of $x$ (particle’s position) and its velocity along $x-axis$ at that value of $x$, I correctly get the value of $v_y$. For example, the particle was at $x$ = $2$ at $t$ = $1$, and its $v_x$ at $t$ = $1$ was $2$ units. Substituting these values I get $v_y$ = $8$ units, which is true, as we can verify from the velocity-time relation that its $v_y$ was $8$ units at $t$ = $1$.

I tried other values of $x$ & $v_x$, or $x$ & $v_y$, and I always seemed to get correct values of $v_y$, and $v_x$. Except for when I tried calculating $v_x$ when the particle was at $x=0$ at $t$ = $0$.

Since $v_x$ = $\frac{v_y}{2x}$, if I put $x$ = $0$ and $v_y$ = $0$, I end up with an indeterminate form $\frac{0}{0}$. ($v_y$ was $0$ at $x$ = $0$)

Shouldn’t the $v_y$ vs $v_x$ relation show that $v_x$ is equal to $2$ units at $x$ = $0$? Actually $v_x$ is constant, and equal to $2$ units at any instant, according to the velocity-time relation.

Am I doing something wrong?